I seek an example that satisfies the following: $\mathbb{P}[A_k > 0 \: \: \: infinitely \: \: often] = 0$ However, $\mathbb{E}[A_k] = ∞$ for all $k$
where $A_k$ for $k>0$ are non-negative, independent R.V.s.
I thought perhaps Borel-Cantelli would be beneficial, but it doesn't seem like it would be.
What about $(B_k)_{k\geq 1}$, $(C_k)_{k\geq 1}$ being respectively a sequence of independent Bernoulli random variables and a sequence of independent standard Cauchy random variables, with $B_k\sim\mathrm{Bern}\left(\frac{1}{k^2}\right)$; and letting $A_k \stackrel{\rm def}{=} B_k \lvert C_k\rvert$ for all $k\geq 0$?
Then: (i) $\mathbb{E}[A_k] = \frac{1}{k^2} \mathbb{E}[\lvert C_k\rvert] = \infty$ for all $k$, but (ii) $\mathbb{P}\{A_k>0\} = \frac{1}{k^2}$ for every $k$.
Applying Borel—Cantelly to $E_k\stackrel{\rm def}{=} \{ A_k > 0\}$ then shows that, since $\sum_{k=1}^\infty \mathbb{P} E_k < \infty$, we have $$\mathbb{P}\{A_k > 0 \text{ i.o.}\} = \mathbb{P}(\lim\sup_k E_k ) = 0.$$